Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in CPP):

Solution:
#include <iostream> 
#include <vector>
#include <algorithm>
#include <iomanip>

using namespace std;

long C[51][51];

int main(int argc, char *argv[]) {
	int N, A, B;
	cin >> N >> A >> B;

	vector<long> v(N);
	for (int i = 0; i < N; i++) cin >> v[i];

	sort(begin(v), end(v), greater<long>());

	for (int n = 0; n <= N; n++) C[n][0] = 1;
	for (int n = 1; n <= N; n++)
		for (int k = 1; k <= n; k++)
			C[n][k] = C[n-1][k-1] + C[n-1][k];

	long sum = 0;
	for (int i = 0; i < A; i++) sum += v[i];
	double avg = (double)sum / A;

	long cnt = 0;
	if (v[0] != v[A - 1]) {
		int n = 0, k = 0;
		for (int i = 0; i < N; i++) {
			if (v[i] != v[A - 1]) continue;
			n++;
			if (i < A) k++;
		}
		cnt = C[n][k];
	} else {
		int n = 0;
		for (int i = 0; i < N; i++)
			if (v[0] == v[i])
				n++;

		for (int k = A; k <= B && k <= n; k++)
			cnt += C[n][k];
	}

	cout << fixed << setprecision(20) << avg << endl;
	cout << cnt << endl;

	return 0;
}